1. for any sets X and Y, we define the symmetric difference X ∆Y by:
X∆Y = (X - Y) ∪ (Y - X) = (X ∪ Y) - (X ∩ Y)
Prove the following:
-for all sets A,B and C, if A - (B ∩ C) = ∅ then A- C ⊆ B
- for all sets A,B and C, if A ⊆ B ∪ C and B ⊆ C ∪ A then A∆B ⊆ C
- for all sets A,B and C, if A∆B = A∆C then B ⊆C
Thanks for your help.
To prove the given statements, we can use set theory laws and logical reasoning. Let's go through each step:
1. Proving A - (B ∩ C) = ∅ → A - C ⊆ B
First, assume that A - (B ∩ C) = ∅, which means that A and (B ∩ C) have no elements in common. To prove A - C ⊆ B, we need to show that if an element x belongs to A - C, then it must also belong to B.
Let's take an arbitrary element x that belongs to A - C. This implies that x belongs to A but not to C. Since x belongs to A and A - (B ∩ C) = ∅, we can conclude that x does not belong to B ∩ C. Therefore, x cannot belong to both B and C.
As x doesn't belong to C, it satisfies the condition A - C ⊆ B, because if x belongs to A - C, it must belong to B since it doesn't belong to C.
Hence, we have proved that A - (B ∩ C) = ∅ → A - C ⊆ B.
2. Proving A ⊆ B ∪ C and B ⊆ C ∪ A → A∆B ⊆ C
Assume that A ⊆ B ∪ C and B ⊆ C ∪ A. We need to show that A∆B ⊆ C, meaning that if an element x belongs to A∆B, then it must also belong to C.
The set A∆B is defined as (A - B) ∪ (B - A), which corresponds to the elements that belong to either A or B but not both.
Let's consider an arbitrary element x that belongs to A∆B. This means that x belongs either to A - B or B - A.
- If x belongs to A - B, it implies that x belongs to A but not to B. By the condition A ⊆ B ∪ C, we know that x must also belong to C, because if it didn't belong to C, it would contradict A ⊆ B ∪ C.
- If x belongs to B - A, it implies that x belongs to B but not to A. By the condition B ⊆ C ∪ A, we can infer that x also belongs to C, as otherwise, it would violate B ⊆ C ∪ A.
In both cases, we conclude that x belongs to C. Therefore, A∆B ⊆ C.
Hence, we have proved that A ⊆ B ∪ C and B ⊆ C ∪ A → A∆B ⊆ C.
3. Proving A∆B = A∆C → B ⊆ C
Assume that A∆B = A∆C. We need to show that B ⊆ C, meaning that every element in B also belongs to C.
Let's take an arbitrary element x that belongs to B. To prove B ⊆ C, we need to show that x also belongs to C.
Since x belongs to B, it can either belong to A - B or B - A.
- If x belongs to A - B, it implies that x belongs to A but not to B. By the definition of symmetric difference, x also belongs to A - C. Therefore, x belongs to A∆C.
- If x belongs to B - A, it implies that x belongs to B but not to A. By the definition of symmetric difference, x also belongs to C - A. Therefore, x belongs to C∆A.
Since A∆B = A∆C, we can say that x belongs to A∆B as well. By the definition of symmetric difference, x belongs to (A ∪ B) - (A ∩ B). This implies x belongs to (A ∪ B) - (C ∩ A), which simplifies to (A ∪ B) - (A ∩ C).
To prove that x belongs to C, it is sufficient to show that x doesn't belong to (A ∪ B) - (A ∩ C).
Assume, for contradiction, that x belongs to (A ∪ B) - (A ∩ C). This means x belongs to (A ∪ B) and x doesn't belong to (A ∩ C). Since x belongs to A∪B, it implies x belongs either to A or B. If x belongs to A, then it must belong to A ∩ C, which contradicts the assumption. If x belongs to B, then it doesn't belong to C, again contradicting the assumption.
Hence, the assumption that x belongs to (A ∪ B) - (A ∩ C) results in a contradiction. Therefore, x doesn't belong to (A ∪ B) - (A ∩ C), and we can conclude that x belongs to C.
As x was an arbitrary element from B, we have shown that every element in B belongs to C. Therefore, B ⊆ C.
Hence, we have proved that A∆B = A∆C → B ⊆ C.
By going through these logical steps, we have successfully proven the given statements using set theory laws.